Question

What is `int ((x^2) / (sqrt(1 - x))) dx`?

Answer 1

`I=-2(1-x)^(1/2)+(4(1-x)^(3/2))/3-(2(1-x)^(5/2))/5+c`

Explanation:

`I=intx^2/sqrt(1-x)dx`
Take, `sqrt(1-x)=t=>1-x=t^2=>x=1-t^2=>dx=-2tdt`
`I=int(1-t^2)^2/cancel(t)(-2*cancel(t)*dt)=-2int(1-2t^2+t^4)dt`
`=>I=-2[t-(2t^3)/3+t^5/5]+c`
`=>I=-2t+(4t^3)/3-(2t^5)/5+c`
`=>I=-2(1-x)^(1/2)+(4(1-x)^(3/2))/3-(2(1-x)^(5/2))/5+c`

Answer 2

The answer is `=-2(1-x)^(1/2)+4/3(1-x)^(3/2)-2/5(1-x)^(5/2)+C`

Explanation:

Perform the substitution

`u=(1-x)`, `=>`, `du=-dx`

`x^2=(1-u)^2`

Therefore,

`I=int(x^2dx)/(sqrt(1-x))=-int((1-u)^2du)/(sqrt(u))`

`=-int((1-2u+u^2)du)/u^(1/2)`

`=-intu^(-1/2)du+2intu^(1/2)du-intu^(3/2)du`

`=-2u^(1/2)+4/3u^(3/2)-2/5u^(5/2)`

`=-2(1-x)^(1/2)+4/3(1-x)^(3/2)-2/5(1-x)^(5/2)+C`